 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-4/3a/e/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For

x \neq = 0

V^{'} (x) = 2 x^{- 3} e^{- 1/ x^{2}}

never vanishes. At

x = 0

, the smooth flat extension has

V^{'} (0) = 0

, and

V (x) > V (0) = 0

for every

x \neq = 0

. Hence $x = 0$ is the unique equilibrium and is stable.

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